In a current-carrying copper wire
with cross section 𝜎 equals 4.0 millimeters squared, the drift velocity is 0.040
centimeters per second. Find the total current running
through the wire. Use a value of 8.49 times 10 to the
28th inverse cubic meters for the number density of copper.
In this wire, which has a
cross-sectional area called 𝜎, there are electrons which move along with a drift
velocity given in the problem statement. This drift velocity, along with the
number density of electrons in this wire, will help us solve for the overall current
that flows through the wire, which we can call 𝐼.
As an equation, we can write that
the current 𝐼 is equal to the number density of electrons in the wire times the
magnitude of the charge on each one of those electrons multiplied by the
cross-sectional area of the wire, given in our problem as 𝜎, times the drift
velocity 𝑣 sub 𝑑.
In the problem statement, we’re
given the number density of electrons as well as their drift velocity and the
cross-sectional area of our wire. We can approximate the magnitude of
the charge on each charge carrier, the electrons, as 1.6 times 10 to the negative
19th coulombs. We’re now ready to plug in and
solve for the current 𝐼 running through this copper wire.
When we do, we’re careful to
replace our cross-sectional area units with meters squared and to express the drift
velocity of the electrons in meters per second. We do these conversions so the
units of these terms are consistent with the rest of the terms in the
expression. Looking over the units in each of
our four terms, we see that the units of meters cubed in the denominator cancels out
with meters squared times meters in the numerator.
Our resulting units of this
multiplication then will be coulombs per second. That is units of amperes. Since we’re solving for a current,
this is a sign that we’re on the right track. To two significant figures, 𝐼 is
22 amperes. That’s the current running through
this copper wire.